You're on a roll — here's the solution for question 3: Question 3: How many chloride ions are present in $1.7 \text{g}$ of magnesium chloride crystals? (Avogadro's constant = $6.0 \times 10^{23}$, Mg = 24, Cl = 35.5) Step 1: Determine the chemical formula of magnesium chloride. Magnesium is a Group 2 element, forming $\text{Mg}^{2+}$ ions. Chlorine is a Group 17 element, forming $\text{Cl}^{-}$ ions. To balance the charges, the formula for magnesium chloride is $\text{MgCl}_2$. Step 2: Calculate the molar mass of magnesium chloride ($\text{MgCl}_2$). Given atomic masses: Mg = 24, Cl = 35.5 $$ \text{Molar mass of MgCl}_2 = \text{Molar mass of Mg} + (2 \times \text{Molar mass of Cl}) $$ $$ \text{Molar mass of MgCl}_2 = 24 \text{ g/mol} + (2 \times 35.5 \text{ g/mol}) $$ $$ \text{Molar mass of MgCl}_2 = 24 \text{ g/mol} + 71 \text{ g/mol} = 95 \text{ g/mol} $$ Step 3: Calculate the moles of magnesium chloride in $1.7 \text{g}$. Given mass of $\text{MgCl}_2 = 1.7 \text{ g}$ $$ \text{Moles of MgCl}_2 = \frac{\text{Mass}}{\text{Molar mass}} $$ $$ \text{Moles of MgCl}_2 = \frac{1.7 \text{ g}}{95 \text{ g/mol}} \approx 0.01789 \text{ mol} $$ Step 4: Calculate the number of $\text{MgCl}_2$ formula units. Using Avogadro's constant ($6.0 \times 10^{23}$ formula units/mol): $$ \text{Number of MgCl}_2 \text{ units} = \text{Moles} \times \text{Avogadro's constant} $$ $$ \text{Number of MgCl}_2 \text{ units} = 0.01789 \text{ mol} \times 6.0 \times 10^{23} \text{ units/mol} $$ $$ \text{Number of MgCl}_2 \text{ units} \approx 0.10734 \times 10^{23} = 1.0734 \times 10^{22} \text{ units} $$ Step 5: Calculate the number of chloride ions. From the formula $\text{MgCl}_2$, each formula unit contains $2$ chloride ions ($\text{Cl}^{-}$). $$ \text{Number of Cl}^{-} \text{ ions} = 2 \times \text{Number of MgCl}_2 \text{ units} $$ $$ \text{Number of Cl}^{-} \text{ ions} = 2 \times 1.0734 \times 10^{22} $$ $$ \text{Number of Cl}^{-} \text{ ions} = 2.1468 \times 10^{22} $$ Rounding to two significant figures (due to $1.7 \text{g}$), we get: $$ \text{Number of Cl}^{-} \text{ ions} = 2.1 \times 10^{22} $$ The number of chloride ions present is $\boxed{\text{2.1} \times 10^{22}}$. What's next? Send 'em.